Scenario A: Assume you can invest \$40,000 each year starting at age 23 and ending at age $n$, earning a 7% return. Assume you leave the money invested until age 60. Scenario B: Assume you can invest \$40,000 each year starting at age $n+1$ and ending at age 60, earning a 7% return. Assume you leave the money invested until age 60. For what age $n$ would the account value at age 60 of Scenario A equal Scenario B?
I have set up the following sums to represent this problem: $$\sum_{k=23}^{n}40000(1.07)^{60-k}=\sum_{k=n+1}^{60}40000(1.07)^{60-k}$$
How do you solve for $n$? You can just try all the numbers and see where they come out closest, but somehow Wolfram Alpha comes up with $n=31.1558$. A fractional number doesn't make since in the context of sum indices, so I have no idea how it's getting this.
Is there some continuous representation of the sum that it's using? How do you calculate $31.1558$?
It's because there doesn't exist a natural indice that satisfies the equation. From above, we have $$40000 \sum_{k=23}^{n} (1.07)^{60-k} = 40000 \sum_{k=n+1}^{60} (1.07)^{60-k}$$ so $\sum_{k=23}^{n} (1.07)^{60-k} = \sum_{k=n+1}^{60} (1.07)^{60-k}$. Using sum of geometric series formula, we have $$\frac{{1.07}^{60-n}(1-1.07^{n-23+1})}{1-1.07} = \frac{{1.07}^{0}(1-1.07^{60-n})}{1-1.07}$$ Solving for n, $1.07^{60-n} - 1.07^{38} = 1-1.07^{60-n} \implies 1.07^{60-n} = (1+ 1.07^{38})/2$, then natural logarithm of both sides gives $(60-n)\ln 1.07 = \ln \frac{1+1.07^{38}}{2} \implies n = 60 - \frac{\ln \frac{1+1.07^{38}}{2}}{\ln 1.07} = 31.1558$.